Calculating a resistor divider circuit to give an output voltage of between 2.5v and 4.5v

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
Hello all - Please help!
This should be so very easy but my head has gone blank.
I am trying to calculate a resistor divider circuit to give an output voltage of between 2.5 volts and 4.5 volts
What I want to use is 2x variable potentiometers (values can be either 2k, 5k, 10k, 20k or 50k) and 1x or more resistors.
I also want the circuit to be as sensitive as possible by varying the voltage output by about 0.1 volts
Power supply is 12vDC

The voltage drop across R1 is to be about 7.5v giving an output of 4.5v when VR1 and VR2 is at 0 ohms.
But I think I need another resistor at node C to ensure that the output does not drop below 2.5v

The schematic below is what I am working with...
How can I lay out the circuit to get the outputs I want?
Thanks.
Resistor Divider1.jpg
 

ericgibbs

Joined Jan 29, 2010
18,849
hi FP,
Why do you need two variable resistors, in those locations.?
You should also allow for the current drawn by the load ie: you show a meter.
E
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
hi FP,
Why do you need two variable resistors, in those locations.?
You should also allow for the current drawn by the load ie: you show a meter.
E
Yes you are probably correct, but I wanted the potentiometers to represent two thermistors. When VR1 is set at a high resistance (thermister cool), I want to vary VR2 (thermister) so that I can determine at what position the VR2 needs to be for the output to be about 3.5 to 3.9 volts.
Susequently depending on the position of VR1 to vary VR2 so the the output is about 3.5 to 3.9 volts.
I don't think I'm putting this very well, but I hope you understand.
 

Dodgydave

Joined Jun 22, 2012
11,302
Yes you are probably correct, but I wanted the potentiometers to represent two thermistors. When VR1 is set at a high resistance (thermister cool), I want to vary VR2 (thermister) so that I can determine at what position the VR2 needs to be for the output to be about 3.5 to 3.9 volts.
Susequently depending on the position of VR1 to vary VR2 so the the output is about 3.5 to 3.9 volts.
I don't think I'm putting this very well, but I hope you understand.
Then you first need to select your Thermistor as this will set the other resistors, thermistors have different resistance at 25C, they vary from 10K, 33K, 47K, 100K. the most common is a 10K at 25C.
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
The 2 thermisters are 10k at 25C
But instead of using hot and cold water to vary the thermisters, I would like to replicate the thermisters by using potentiometers and marking the positions of the potentiometers in relation to the other so that the output is somewhere around 3.5 to 3.9 volts
Additional..
VR1 (Thermister) is to be between 0C and 25C - 32.7k down to 10k
VR2 (Thermister) is to be between 15C and 60C - 15.7k down to 2.5k
 
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Dodgydave

Joined Jun 22, 2012
11,302
The 2 thermisters are 10k at 25C
But instead of using hot and cold water to vary the thermisters, I would like to replicate the thermisters by using potentiometers and marking the positions of the potentiometers in relation to the other so that the output is somewhere around 3.5 to 3.9 volts
Additional..
VR1 (Thermister) is to be beween 0C and 25C - 32.7k down to 10k
VR2 (Thermister) is to be beween 15C and 60C - 15.7k down to 2.5k
Then you've just answered your own question!
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
Thanks all.
I've been doing some calculations and given the supply will be 12v, I think that using
R1=10k
VR1=20k and
VR2= 20k
this way I can at least get an outputs of between 0 and 8 volts
ie. if R1=10k, VR1=20k, VR2=0 then output will be 0 volts
and
if R1=10k, VR1=0, VR2=20k then output will be 8 volts
and
if R1=10k, VR1=10k, VR2=10k then output will be 4 volts
----------------------------------------------------------------------------------------
so varying VR1 and VR2 I can get the voltages of between 2.5 volts and 4.5 volts.
....but I think the sensitivity will not be very good.
I may try to do some more, different calculations.
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
hi FP,
What is the purpose of this variable voltage divider.?
What is the output going to be used for.??

E
Hi ericgibbs,
To check the operation of a circuit that is attached to the output.
The circuit operates on an input voltage of between about 3.5v and 4.0v
The voltage divider is merely to replicate the thermisters operation.
 

bertus

Joined Apr 5, 2008
22,277
Hello,
Additional..
VR1 (Thermister) is to be beween 0C and 25C - 32.7k down to 10k
VR2 (Thermister) is to be beween 15C and 60C - 15.7k down to 2.5k
What is the relation between the Thermisters?
Do the follow eachother or is there an other relation?

Bertus
 

ericgibbs

Joined Jan 29, 2010
18,849
To check the operation of a circuit that is attached to the output.
hi FP,
I guess you realise as soon as you connect a moderate resistive load to that divider, your calculations are no longer correct.?

I would suggest you post a proposed circuit with a simple explanation of its purpose, in that way we can help you.

E
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
hi FP,
I guess you realise as soon as you connect a moderate resistive load to that divider, your calculations are no longer correct.?
I would suggest you post a proposed circuit with a simple explanation of its purpose, in that way we can help you.
E
This is the circuit that I have made and works well.

But I like to test the circuit by adding a voltage divider in place of the thermisters just so that I do not have to keep heating up cups of water all the time.
At the non-inverting input the voltage is about 3.77 to 3.84 volts.
The resistor divider is to replace the thermisters to adjust the inverting input from 2.5v and up to about 4.5v
As long as I know the approximate positions of the potentiometers I can work out whether the circuit is working correctly or not.
I have made a voltage divider as in post 1 with R1=10k VR1=20k and VR2=20k but I would just like it a bit more sensitive.
EDIT: There is a 100nF decoupling capacitor between the VCC and GND of the opamp but not shown on the schematic.
 

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ericgibbs

Joined Jan 29, 2010
18,849
hi FP,
Thats much more helpful.
I assume when you say Solar Panel, its not a PV but a solar heated water radiator and the thermistor is clamped to the radiator or pipe work.?
From my calculations the higher the radiator water temp the higher the pool temperature has to rise to switch the circulating pump OFF.
Assuming the solar water heater can only reach say a 50C maximum , the pool water could be close to 30C.

If the pool water temp is say 15C, the solar heater will have to be > 30C before pump switch ON,

You are not getting the maximum heat exchange control with that simple circuit,

Eric
 

Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
hi FP,
Thats much more helpful.
I assume when you say Solar Panel, its not a PV but a solar heated water radiator and the thermistor is clamped to the radiator or pipe work.?

From my calculations the higher the radiator water temp the higher the pool temperature has to rise to switch the circulating pump OFF.
Assuming the solar water heater can only reach say a 50C maximum , the pool water could be close to 30C.
If the pool water temp is say 15C, the solar heater will have to be > 30C before pump switch ON,
You are not getting the maximum heat exchange control with that simple circuit,

Eric
Hi ericgibbs,
Yes,the solar panel is/will be a solar heated water radiator and 'hot' thermistor is clamped to the radiator or pipe work. The other 'cool' thermister measures the pool water temperature.
In theory I believe the solar panel could reach temperatures of possibly about +70C to +80C in hot climates (Mediterranean areas) when the sun is shining which could yield an output water temperature of maybe +40C to +50C.
In cooler climates (Eg. North of London, England) the solar panel would perhaps reach temperatures of possibly about +50C to +60C when the sun is shining which could yield an output water temperature of maybe +30C to +40C - and maybe even higher
I have designed the circuit so that the water pump will switch on if the solar panel water is approx. +20C above that of the pool water temperature. So if the water temperature is +18C the solar panel water would need to be about +38C (18+20) and the pump switches on.

© My design.
 
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ericgibbs

Joined Jan 29, 2010
18,849
hi FP,
At least our 'sums' for the project temp's are close.
I would consider adding an upper pool temp limit for the pump operation.

Ideally what would you like the specification for the 'calibration' of the unit to achieve.?
Would it be for production testing/pre-setting only or be a part of the unit for the user to set to their preferred limits.?

E
BTW: I no longer design/manufacture products, so no worry about me copying.;)

EDIT:
The Thermistor parameters I am using are.
R0 = 10k
T0 = 25C
B = 3988
TK = 273.15
 
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Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
Hi ericgibbs,
After much experimentation and testing, I decided that the optimum differential temperature would be/is about +20C. I can however alter this very easily by changing the 4.7k resistor to another value.
With the 10k and 4.7k voltage divider this gives the non-inverting input of the opamp about 3.77 to 3.84 volts
So, it will be used for production testing, I would want a very simple way of checking, testing and the circuit that it is operating at the correct thesholds before I ship them out to customers and making sure it is accurate.
I am currently using a 10k, and 2x 20k potentiometers as in my post 8 above, but I would like something just a bit more sensitive.
I was hoping that I could make something easy and simple circuit so that I can get outputs of between 2.5 and 4.5 volts and incremented by about 0.1 volts - but perhaps 2.5 is too low and 4.5 is a bit too high...Maybe something like 3.5 and 4.2 volts would be ideal.
 
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ericgibbs

Joined Jan 29, 2010
18,849
hi,
Lets say 3.5V thru 4.5v
Using a 1000R , 10 or 20 turn linear pot [ optional mech dial]
For 1v drop across the pot, the series current is 1mA.
Using a 12V supply and assuming 1mA. thats 12K total.

12k - 1k[pot] leaves 11k for the fixed resistors.

Top resistor is 7500R
Pot 1k
Bottom resistor 3500R
Gives 3.5V thru 4.5v at the pot wiper for full rotation

E

Check this link for indicator knobs.
https://uk.rs-online.com/web/c/?sra=oss&r=t&searchTerm=potenetiometer+knobs

EDIT:

Added a sim.
 

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Thread Starter

Ford Prefect

Joined Jun 14, 2010
245
HI ericgibbs,

YES!! this is very good. thank you ericgibbs :)
I cannot get a 3.5k ohm resistor at my local shop they only sell 3.6k ohm (+/- 5%).
with the 3.6k I am calculating a voltage output of between 3.57v and 4.56v - this I think will be close enough.
DiffController-2.jpg

I may just juggle about with the resisitors values and see if I can get it even closer :)
 
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